31. Diffraction: a few important illustrations

Babinet’s Principle

Diffraction gratings

X-ray diffraction: Bragg scatteringand crystal structures

A lens transforms a Fresnel diffraction problem into a Fraunhofer diffraction problem

Diffraction and image resolution: the Rayleigh criterion

Diffraction regimes

Near field: z D or less

Fresnel regime: z >> D (paraxial)

Fraunhofer regime: z >> pD2/l and z >> D

D

input plane wave

aperture

z

observation plane

l

In the Fraunhofer regime, the diffraction pattern is a Fourier transform.

1

2

3

Note: this is the typical ordering of the

regimes; if the aperture is smaller

than the wavelength, then regimes 2 and 3

would be swapped.

Babinet’s Principle

0 0 0 1 0 1 1 1 1 1, exp , A

jkE x y x x y y Aperture x y dx dy

z

Fraunhofer diffraction is a Fourier transform:

Consequence: A complementary aperture (one which is the inverse of the original one) must give the same diffraction pattern:

0 0 0 1 0 1 1 1 1 1, exp 1 ( , ) C

jkE x y x x y y Aperture x y dx dy

z

Jacques Babinet1794-1872

0 0 0 0, , C AE x y E x yThus: at all points except x0 = y0 = 0

Babinet’s Principle: example

an array of holes

an array of anti-holes

The Diffraction GratingA diffraction grating is a slabwith a periodic modulationof any sort on one surface.

The modulation can be intransmission, reflection, orthe phase delay of a beam.

The grating is then said to bea transmission grating,reflection grating, orphase grating, respectively.

What happens when a plane wave illuminates an object of this sort?

A sinusoidal modulation

A square modulation

Diffraction Grating Mathematics

1 1 1( , ) cos(2 / )Aperture x y A B x ap

0 1 0 1 1( ) cos(2 / ) exp ( ) jk

E x A B x a x x dxz

p

This is just the Fourier transform of a constant (A) plus a cosine Bcos(2px1/a):

x0 = zl / a x0 = zl / ax0 = 0

first order negative first orderzeroth order

a

0 00([ / ] 2 / 2) ([ / ] 2 / )([ / ]) kB kx z a B kx ax z zp p p p p

As an example, consider a sinusoidal modulation of the transmission:

where a is the "grating spacing."

0 1 0 1 1( ) exp ( )jk

E x Aperture x x x dxz

The Fraunhofer diffracted field is:

(this is a one-dimensional aperture function, so we ignore the y1 integral)

Diffraction orders

0 /x z al

z

x0

0 0x

0 /x z al

1 0 / /x z a l

Because x0 depends on l for the +1 and 1 orders, different wavelengths are separated.

The longer the wavelength, the larger its diffraction angle.(except for zeroth order)

/z al

No wavelength dependence in zero order.

input plane wave

Diffraction Grating Mathematics:Higher OrdersWhat if the periodic modulation of the transmission is not sinusoidal?

1 1 0 1 1 2 1 3 1( , ) cos(2 / ) cos(4 / ) cos(6 / ) ... Aperture x y A A x a A x a A x ap p p

Since it's periodic, we can use a Fourier Series for it:

Keeping up to third order, the resulting Fourier Transform is:

3 0

2 0

1 0

0 0

1 0

2 0

3 0

([ / ] 6 / )

([ / ] 4 / )

([ / ] 2 / )

2 ([ / ])

([ / ] 2 / )

([ / ] 4 / )

([ / ] 6 / )

A kx z a

A kx z a

A kx z a

A kx z

A kx z a

A kx z a

A kx z a

p pp pp pp p pp pp p

A square modulation is commonly used. It has many orders.

0

0

0

0

0

6 / / 3rd order

4 / / 2nd order

2 / / 1st order

0 0th order

3

2

1

2 / / -1 1st

x z ka z a

x z ka z a

x z ka z a

x

x z ka z a

p lp lp l

p l

0

0

order

4 / / -2nd order

6 / / -3rd orde3 r

2

x z ka z a

x z ka z a

p lp l

The Grating Equation

An order of a diffraction grating occurs if:

where m is an integer.

0[ / ] or

sin( )

l lm

a x z m

a m

sin( ) sin( )m ia m l

Remember that the diffracted angle can be negative, too. But it cannot be larger than ±90°.

This equation assumed normal incidence and a small diffraction angle, however. One can derive a more general result, the "grating equation," using a tilted input beam:

Blazed Diffraction Grating

By tilting the facets of the grating so the desired diffraction ordercoincides with the specular reflection from the facets, the gratingefficiency can be increased.

“Specular” means angle of incidence equals angle of reflection.

Input beam

Even though both diffracted beams satisfy the grating equation, one is vastly more intense than the other.

Efficient diffraction

surface normal

Inefficient diffraction

Diffraction from a periodic array: Bragg’s Law

We can also derive the grating condition by looking at the path length difference between waves scattered from adjacent sites in a periodic array.

This leads us to Bragg’s Law, for scattering of x-rays from crystalline materials. This is essentially equivalent to the grating diffraction problem.

A periodic array of atoms in a lattice is illuminated by a plane wave. Each atom produces a scattered spherical wave.

At a certain specific angle, the scattered waves interfere constructively, just as in the case of waves scattered from a grating.

X-ray Crystallography

The tendency of diffraction to expand the smallest structure into the largest pattern is the key to the technique of x-ray crystallography, in which x-rays diffract off the nuclei of crystals, and the diffraction pattern reveals the crystal molecular structure.

This is the standard method for determining the crystal structure of any solid. X-ray diffraction pattern from

polycrystalline (left) and single-crystal (right) Cr2O3

crystal structure of Cr2O3

Application: Finding the structures of biomolecules Knowing where all the atoms are in a large

molecule is difficult. But it is crucial for understanding how molecules interact and function.

If you can persuade the molecules to form a crystal, that is, to arrange themselves regularly in space, then you can use x-ray diffraction to find the location of every atom.Fourier

transform

This idea is most important in biology.

the three-dimensional structure of penicillin

red = oxygenblue = nitrogenyellow = sulphurgreen = carbonwhite = hydrogen

Dorothy Crowfoot Hodgkin1910 - 1994

The most important x-ray diffraction pattern in history

Data obtained by Rosalind Franklin, 1952.

Rosalind Franklin1920 - 1958

Nobel prize obtained by Crick and Watson, 1962

Francis Crick1916 - 2004

James Watson1928 -

Diffraction treatment of a spherical lens

An ideal lens has unity transmission, but it introduces a phase delay in proportion to its thickness at a given point (x1,y1):

where D(x1,y1) is the thickness at the point (x1,y1).

1 1 1 1( , ) exp[ ( 1) ( , )] Dlens x y j n k x y

2 2 21 1 1 1 1( , ) exp ( 1) ( )lens x y j n k R x y d

where we have neglected constant (independent of x1, y1) phase delays.

2 21 1 1 1 1( , ) exp[ ( 1) 2 ] lens x y jk n x y R

2 22 2 2 2 2 2 1 1

1 1 1 1 1 1 1 11

1 ( ) /2

x yR x y R x y R R

R

plane wave illumination

lens circle, radius R1

d

2 2 21( , )x y R x y dD

Compute D(x1,y1):

In the focal plane, the diffraction problem is Fraunhofer

A lens is an aperture function with unity magnitude, but with a phase delay proportional to its thickness at a given point (x1,y1):

2 21 1 1 1 1 1

1

1, , exp

2

n kaperture x y lens x y j x y

R

1 1( 1)( / 2 ) ( / 2 ) or 1/ ( 1)(1/ )n k R k z z n R

the quadratic terms x12 and y12 will cancel provided that:

But this is the Lensmaker’s Formula! The distance z which satisfies this condition is the focal length of the lens!

2 2

0 1 0 1 1 10 0 1 1 1 1

( 2 2 ( )( , ) exp ,

2 2

x x y y x yE x y jk lens x y dx dy

z z

If we substitute this result into the Fresnel (not Fraunhofer) integral:

A lens brings the far field in to its focal plane

If we look in a plane one focal length beyond a lens, we are in the Fraunhofer regime, even if it isn’t far away! So we see the Fourier Transform of any object that is in front of the lens.

A lens in this configuration is said to be a “Fourier-Transforming lens.”

Object: an opaque screen with a square hole

Focal plane: sinc(x) * sinc(y)

A focused Gaussian beam produces a Gaussian spot at the focus. But a focused plane wave produces an Airy pattern.

input: I(x1,y1) ~ Gaussianin the Fourier plane: still Gaussian, but narrower: rspot depends on winput

0 0 1 1, ,Gaussian x y FT Gaussian x y

input: I(x1,y1) ~ constant

in the Fourier plane: an Airy pattern [J1(r)/r2

0 0 1 1, , , for lensAiry x y FT const x y Rr

The size of the Airy pattern is determined by the size of the lens.

(plane wave)

Application: resolution limit of a telescope

Consider a telescope looking at a distant star…

Fourier transform plane

In this case, the telescope pupil (the size of the lens) is the limiting aperture. A plane wave illuminating this circular aperture produces an Airy pattern in the image plane. The size of this image is determined by the size of the lens, D, and the focal length, f, according to:

1.22 1.22 /# spot

ff

D

lr l

The angular spread D is given by rspot/f, and therefore:

1.22D D l

an Airy pattern!D

Application: resolution limit of a telescope

Now consider a telescope looking at two distant stars…

Fourier transform plane

Rayleigh criterion: it is possible to resolve the two stars when the peak of the Airy pattern of one coincides with the first minimum of the Airy pattern of the other one.

Thus, the telescope is able to resolve two stars if they are separated in the sky by at least an angle of:

min 1.22D D lResolution can be improved by (a) decreasing l, or (b) increasing D.

Can we tell them apart? Are the twostars resolved?

The Rayleigh criterion

Once again, Rayleigh is The Man…

John William Strutt, 3rd Baron Rayleigh

1842 - 1919

“This rule is convenient on account of its simplicity and it is sufficiently accurate in view of the necessary uncertainty as to what exactly is meant by resolution.”

-4 -3 -2 -1 0 1 2 3 4

The Rayleigh criterion

These two Airy patterns are just barely resolved, according to Rayleigh’s criterion.

The Rayleigh criterion is a somewhat arbitrary definition of resolution. There are other criteria that are also often used.

‘valley to peak ratio’ = 73% when the Rayleigh criterion is satisfied.

e.g., see the “Sparrow Criterion”

Application: resolution limit of a telescope

Hubble space telescope

In fact, Hubble usually achieves a resolution of about 0.1 arc-sec, which is about two times the diffraction limit. This is limited by spherical abberation of the focusing mirror.

2.4 metersD

So, for green light,

500 nanometersl

the resolving power of Hubble is:

7min 2.5 10 radians

0.05 arc-seconds

D

The size of the reflecting mirror in Hubble is: